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    (Original post by Hai_Ann)
    Can anyone help me solve this?
    I thought it was some sort of 3D parallelogram at first.
    So for the area I got... 100x^2 y+16x^2+2xy
    I don't think that's right and I don't know what to do next....
    Hey there, I would recommend learning the trignometric ratios for triangles with the angles ( 30, 60 and 90 ) as well ( 45, 45 and 90 ). They come in handy when tackling such questions..

    Alright so, assuming you've already found y, which is 100 / x^2. To find the area of metal used, you're basically finding the area of all faces added together.

    You have two trapezium faces, 1 rectangle base, and two inclined rectangles.

    The area of the trapezium faces => 2 [ 1/2 * ( 8x + 10x ) * x ] = 18x^2

    The area of the base => 8x * y => 8x * 100 / x^2 => 800 / x

    Now for the last bit, they tell you in the question, the two smaller rectangles are inclined at 45 degrees to the horizontal. With this in mind, you end up with:

    The ratios in this triangle are, 1:1:√2, with the 1's belonging to the adjacent and opposite, and √2 for the hypotenuse.

    Since x is an opposite, and we're trying to find the hypotenuse ......
    that leaves us with x : ? ---> 1 : √2 giving us ? = x√2

    The area of the smaller rectangles : 2 (x√2 * y) => 2( x√2 * 100 / x^2 )
    = 200x√2 / x^2 = 200√2 / x

    Area of metal used => 18x^2 + (800/x) + ( 200√2 / x )

    You factorize the two which have x in their denominator to:

    18x^2 + 200( 4 + √2 ) / x

    Hope any of that made sense....
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    (Original post by Skygon)
    Hey there, I would recommend learning the trignometric ratios for triangles with the angles ( 30, 60 and 90 ) as well ( 45, 45 and 90 ). They come in handy when tackling such questions..

    Alright so, assuming you've already found y, which is 100 / x^2. To find the area of metal used, you're basically finding the area of all faces added together.

    You have two trapezium faces, 1 rectangle base, and two inclined rectangles.

    The area of the trapezium faces => 2 [ 1/2 * ( 8x + 10x ) * x ] = 18x^2

    The area of the base => 8x * y => 8x * 100 / x^2 => 800 / x

    Now for the last bit, they tell you in the question, the two smaller rectangles are inclined at 45 degrees to the horizontal. With this in mind, you end up with:

    The ratios in this triangle are, 1:1:√2, with the 1's belonging to the adjacent and opposite, and √2 for the hypotenuse.

    Since x is an opposite, and we're trying to find the hypotenuse ......
    that leaves us with x : ? ---> 1 : √2 giving us ? = x√2

    The area of the smaller rectangles : 2 (x√2 * y) => 2( x√2 * 100 / x^2 )
    = 200x√2 / x^2 = 200√2 / x

    Area of metal used => 18x^2 + (800/x) + ( 200√2 / x )

    You factorize the two which have x in their denominator to:

    18x^2 + 200( 4 + √2 ) / x

    Hope any of that made sense....
    Nice drawing
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    (Original post by fire_and_ice)
    Nice drawing
    10/10, if this question came up in the exam I would cry.
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    (Original post by fire_and_ice)
    Nice drawing
    I try :banana:
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    Could I have some help with this question please, from the 2014(r) paper

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    (Original post by usmanzubair)
    Make a physics unit 1 thread pls lol we have that before c2
    I've just made one!!!! Please visit
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    (Original post by TheRaspberry)
    Could I have some help with this question please, from the 2014(r) paper

    Try rearranging it by multiplying both sides by 4 sin(2theta) -1. You should end up with sin (2theta) =.
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    (Original post by TheRaspberry)
    Could I have some help with this question please, from the 2014(r) paper

    sin2theta = 4sin2theta - 1
    3sin2theta = 1
    sinINVERSE of 1/3 = 2theta
    you'll get like 19.5
    so 19.5 = 2theta
    19.5 + 180(n)
    and (180-19.5) + 180(n)

    DIVIDE by 2 as its 2 theta
    and identify the values which are in the range.

    *much quicker on paper btw XD*
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    (Original post by TheRaspberry)
    Could I have some help with this question please, from the 2014(r) paper

    Multiply through and you'll end up with sin 2 theta = 4 sin 2 theta - 1

    Re-arrange it to, 4 sin 2 theta - sin 2 theta = 1

    3 sin 2 theta = 1
    sin 2 theta = 1 /3

    Since it's 2 theta, don't forget to double the range as well, from
    [ 0 < theta < 180 ] to [ 0 < theta < 360 ]

    Now solve for 2 theta, get your angles, and then divide them by 2 at the end so that they fit the given range.
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    Could anyone please guide me through an appropriate way to handle such questions? I get how the trapezium rule works but I have no idea how to set up the range of x values.



    Thanks in advance!

    Edit:- Only need help with part (b) and (c).
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    Tip:

    If you have values for theta or x or whatever the unknown constant is, substitute them back into the equation and make sure they ALL work. Also, look at the boundaries. If something is 0 =< x < 360, it is in degress. If something is 0 =< x < 2pi, it is in radians.
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    Thanks all, understand now Much appreciated!
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    so whos got the leaks for c2
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    Bad 😨
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    (Original post by edward090)
    so whos got the leaks for c2
    according to people who tried to open it it's a fake link and doesnt work (some pdf on the internet) and according to some threads it HAS actually leaked however something tells me edexcel will just give everyone the normal paper.
    so yh i wouldn't bother with that
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    (Original post by Skygon)
    Could anyone please guide me through an appropriate way to handle such questions? I get how the trapezium rule works but I have no idea how to set up the range of x values.



    Thanks in advance!

    Edit:- Only need help with part (b) and (c).
    So as not to panic anyone else, this question is not written in the style of any past edexcel question and uses vocabulary that past papers don't use.

    You need to use the trapezium rule using only 3 equally spaced values for x (ordinates) so only 2 strips i.e. 0, pi/6 and pi/3. Work out the corresponding y values using y=cos^2(x). Plug the results into the trapezium rule.

    For part c remember that sin^2(x)=1-cos^2(x) so the answer to c is the area above the curve but under the line y=1

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    (Original post by AJC1997)
    Me, Econ 2 Tuesday, History 2 and Maths C2 on Wednesday, Biology unit 1 on Thursday.

    Why you so mean, Edexcel?
    You doing aqa economics, i feel your pain, i have econ, followed by maths, bio, chem

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    (Original post by gdunne42)
    So as not to panic anyone else, this question is not written in the style of any past edexcel question and uses vocabulary that past papers don't use.

    You need to use the trapezium rule using only 3 equally spaced values for x (ordinates) so only 2 strips i.e. 0, pi/6 and pi/3. Work out the corresponding y values using y=cos^2(x). Plug the results into the trapezium rule.

    For part c remember that sin^2(x)=1-cos^2(x) so the answer to c is the area above the curve but under the line y=1

    Posted from TSR Mobile
    Can you also say that ∫ sin^2x=∫ 1-cos^2x
    Which equals ∫ 1 - ∫ cos^2x which you know from the part before


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    C2 will be easy I just performed my senior recital it was sick Im a flute pro
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    (Original post by Medicjug)
    Can you also say that ∫ sin^2x=∫ 1-cos^2x
    Which equals ∫ 1 - ∫ cos^2x which you know from the part before

    Posted from TSR Mobile
    Yes you can
    That kind of thinking has appeared in recent international papers but not as far as I can remember on a UK paper.......... Yet

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